Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a (quasi-)projective, nonsingular, complex variety. Denote by $\mathcal{T}_X$ its tangent sheaf and by $X^{\mathrm{an}}$ its analytification. I am looking for a proof for the equality

        $\displaystyle \int_X c_n(\mathcal{T}_X) = \chi(X^{\mathrm{an}})$,

i.e. the degree of the top chern class is equal to the topological Euler characteristic of $X$. There's Example 3.2.13 in Fulton's book on intersection theory which briefly mentions this, but it does not give a reference. Can someone help me out with one? Thanks in advance.

share|improve this question
1  
I think the difficulty you're having finding a reference is you're asking the question in too local a setting -- Chern classes are for complex bundles over spaces. In that setting, Wikipedia gives you the reference: en.wikipedia.org/wiki/Chern_class –  Ryan Budney Aug 23 '11 at 0:22
    
You mean the original paper by Chern? I thought that since 1945, it might have been elaborated in a more algebraic setting and greater detail, which is what I was rather looking for. –  Jesko Hüttenhain Aug 23 '11 at 0:51
    
The top Chern class is the obstruction to a section of the bundle. In the case of the tangent bundle, the Poincare-Hopf Index Theorem (see for example Milnor and Stasheff, or the textbook Guillemin and Pollack) tells you evaluation of this class is Euler characteristic. –  Ryan Budney Aug 23 '11 at 0:57
1  
A proof using Riemann--Roch and a Koszul-type resolution of the Diagonal can be found in the notes math.uni-bonn.de/people/hartmann/intersection_theory.pdf on my webpage. –  Heinrich Hartmann Aug 23 '11 at 10:57
    
@Heinrich: That's just perfect! I should come to you all the time with this kind of stuff. Seriously though, thanks a bunch. –  Jesko Hüttenhain Aug 25 '11 at 9:07

2 Answers 2

up vote 11 down vote accepted

As an alternative to R. Budney's answer, one might also notice that the Gauss-Bonnet formula (the one you mention - mind that you must assume that $X$ is projective, otherwise the integral might not even make sense) is a consequence of the Hirzebruch-Riemann-Roch theorem. Indeed, the HRR theorem says $$ \chi(V)=\int_{X}{\rm Td}({\rm T}X){\rm ch}(V) $$ where $$\chi(V):=\sum_{l}{(-1)}^l{\rm rk}(H^l(X,V))$$ is the Euler characteristic of coherent sheaves. Now there is an universal identity of Chern classes $$ {\rm ch}(\sum_{r}(-1)^r\Omega_X^r){\rm Td}(\Omega^\vee_X)=c^{\rm top}(\Omega^\vee_X) $$ (called the Borel-Serre identity). Here $\Omega_X$ is the sheaf of differential of $X$ and thus $\Omega^\vee_X={\rm T}X$. Plugging the element $\sum_{r}(-1)^r\Omega_{X}^r$ into the HRR theorem, one gets $$ \sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\int_{X}c^{\rm top}(TX) $$ and by the Hodge decomposition theorem $$ \sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C})) $$ where $H^r(X({\bf C}),{\bf C})$ is the $r$-th singular cohomology group. The quantity $\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$ is the topological Euler characteristic, so this proves what you want. The HRR theorem is proved in chap. 15 of Fulton's book (or in Hirzebruch's book "Topological methods...") and the Borel-Serre identity is Ex. 3.2.5, p. 57 of the same book.

share|improve this answer
    
Do you use Poincare-Hopf in the proof of the general Gauss-Bonnet theorem? For example, there are common proofs of Gauss-Bonnet in the 2-dimensional (real) manifold case that use something that is very close to the Poincare-Hopf theorem -- the fact that the tangent vector to a closed curve in the plane has total winding number $\pm 1$. –  Ryan Budney Aug 24 '11 at 3:31
    
Another point of view is to use the (related) Lefschetz trace formula. The proof goes as follows: i) The Euler characterstic is the trace of the identity, hence is the intersection of the diagonal with itself in X x X. ii) The intersection of the diagonal with itself is computed by the excess intersection-formula, which gives the intersection in terms of the Chern class of the normal bundle of the diagonal embedding. iii) This is almost by definition the tangent or cotangent-bundle. –  D. Eriksson Aug 12 at 11:12

I'll convert my comments to an answer.

Characteristic classes were originally defined as obstruction classes, going back to the work of Whitney and Stiefel. The top Chern class of a complex bundle is the obstruction to an everywhere non-zero vector field, and frequently called the Euler class.

The reason for the name "Euler class" comes from the case where you're looking at the tangent bundle. When you pair this class against the fundamental class of the manifold, via Poincare duality you can interpret this as the signed intersection number of the 0-section of the tangent bundle with (a transverse perturbation of) itself. Since this number does not depend on the perturbation, there are various ways to compute it and check it's the classical Euler characteristic of the base manifold. I like to use a triangulation, and adapting a vector field to it so that the zeros of the vector field correspond to the (barycentres) of the cells of the triangulation.

The core geometric argument here is the Poincare-Hopf index theorem.

Perhaps there are more adapted ways to get the result in the context you like (I don't know), but I think this is the historical route to the result.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.